Question: Consider the polar curve $r=2\tan(\theta)$ over the interval $\left( 0,2\pi \right)$. At which value of $\theta$ does the graph of $r$ have a vertical tangent line? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\pi}{2}$ only (Choice B) B $\dfrac{3\pi}{2}$ only (Choice C) C $\dfrac{\pi}{2}$ or $\dfrac{3\pi}{2}$ (Choice D) D The graph of $r$ has no vertical tangents.
Explanation: A vertical line has an undefined slope. The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Once we have an expression for the slope of the tangent line, we can look for the $\theta$ -values that make $\dfrac{dx}{d\theta}=0$ but don't make $\dfrac{dy}{d\theta}=0$ (because then $\dfrac{dy}{dx}$ will be indeterminate). For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={2\tan(\theta)}\cos(\theta) \\\\ &=2\sin(\theta) \\\\ y&={2\tan(\theta)} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$. $\begin{aligned} y(\theta)&=2\tan(\theta)\sin(\theta) \\\\ \dfrac{dy}{d\theta}&=2\sec^2(\theta)\sin(\theta)+2\tan(\theta)\cos(\theta) \\\\ &=\dfrac{2\sin(\theta)}{\cos^2(\theta)}+2\sin(\theta) \\\\ \\\\ x(\theta)&=2\sin(\theta) \\\\ \dfrac{dx}{d\theta}&=2\cos(\theta) \end{aligned}$ Now let's solve $\dfrac{dx}{d\theta}=0$ on the interval $\left( 0,2\pi \right)$. $\begin{aligned} \dfrac{dx}{d\theta}&=0 \\\\ 2\cos(\theta)&=0 \\\\ \cos(\theta)&=0 \end{aligned}$ Within our interval, our possible solutions are $\theta=\dfrac{\pi}{2}$ or $\theta=\dfrac{3\pi}{2}$. Notice, however, that $r(\theta)=2\tan(\theta)$ isn't defined at $\theta=\dfrac{\pi}{2}$ or $\theta=\dfrac{3\pi}{2}$. If it's undefined at these points, it definitely doesn't have a tangent there. The graph of $r$ has no vertical tangents. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${0}$ ${\frac{1}{6}\pi}$ ${\frac{1}{3}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{2}{3}\pi}$ ${\frac{5}{6}\pi}$ ${\pi}$ ${\frac{7}{6}\pi}$ ${\frac{4}{3}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{5}{3}\pi}$ ${\frac{11}{6}\pi}$